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Logic!

Fall 2002

CMSC 203 - Discrete

1

Logic
• Crucial for mathematical reasoning
• Used for designing electronic circuitry
• Logic is a system based on propositions.
• A proposition is a statement that is either
true or false (not both).
• We say that the truth value of a proposition
is either true (T) or false (F).
• Corresponds to 1 and 0 in digital circuits
Fall 2002

CMSC 203 - Discrete

2

The Statement/Proposition Game
“Elephants are bigger than mice.”
Is this a statement?

yes

Is this a proposition?

yes

What is the truth value
of the proposition?

true

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3

“520 < 111”

yes


yes

of the proposition?

false

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4

“y > 5”

yes


no

Its truth value depends on the value of y,
but this value is not specified.
We call this type of statement a
propositional function or open sentence.
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5

“Today is January 1 and 99 < 5.”

yes


yes

of the proposition?

false

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6

“Please do not fall asleep.”

no

It’s a request.

no

Only statements can be propositions.
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7

“If elephants were red,
they could hide in cherry trees.”

yes


yes

of the proposition?

probably false

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8

“x < y if and only if y > x.”
yes
yes
… because its truth value
does not depend on
specific values of x and y.
of the proposition?
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CMSC 203 - Discrete

true
9

Combining Propositions
As we have seen in the previous examples,
one or more propositions can be combined
to form a single compound proposition.
We formalize this by denoting propositions
with letters such as p, q, r, s, and
introducing several logical operators.
Fall 2002

CMSC 203 - Discrete

10

Logical Operators (Connectives)
We will examine the following logical operators:
•
•
•
•
•
•

Negation
Conjunction
Disjunction
Exclusive or
Implication
Biconditional

(NOT)
(AND)
(OR)
(XOR)
(if – then)
(if and only if)

Truth tables can be used to show how these
operators can combine propositions to
compound propositions.
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11

Negation (NOT)
Unary Operator, Symbol: ¬
P
true (T)

false (F)

false (F)
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¬P
true (T)

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Conjunction (AND)
Binary Operator, Symbol: ∧
P
T

P∧Q
T

T

F

F

F

T

F

F
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Q
T

F

F

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Disjunction (OR)
Binary Operator, Symbol: ∨
P
T

P∨ Q
T

T

F

T

F

T

T

F
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Q
T

F

F

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Exclusive Or (XOR)
Binary Operator, Symbol: ⊕
P
T

P⊕ Q
F

T

F

T

F

T

T

F
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Q
T

F

F

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Implication (if - then)
Binary Operator, Symbol: →
P
T

P→ Q
T

T

F

F

F

T

T

F
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Q
T

F

T

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Biconditional (if and only if)
Binary Operator, Symbol: ↔
P
T

P↔ Q
T

T

F

F

F

T

F

F
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Q
T

F

T

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Statements and Operators
Statements and operators can be combined in any
way to form new statements.

P
T

Q
T

¬P
F

T

F

F

T

T

F

T

T

F

T

F

F

T

T

T

CMSC 203 - Discrete

18

Fall 2002

¬Q (¬P)∨(¬Q)
F
F

Statements and Operations
Statements and operators can be combined in any
way to form new statements.

P
T

Q
T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

T

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P∧Q ¬ (P∧Q) (¬P)∨(¬Q)
T
F
F

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Equivalent Statements
P

Q

T
T
F
F

T
F
T
F

¬(P∧Q)↔ (¬P)∨(¬Q
¬(P∧Q) (¬P)∨(¬Q)
)

F
T
T
T

F
T
T
T

T
T
T
T

The statements ¬(P∧Q) and (¬P) ∨ (¬Q) are logically
equivalent, since ¬(P∧Q) ↔ (¬P) ∨ (¬Q) is always true.

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Tautologies and Contradictions
A tautology is a statement that is always true.
Examples:
• R∨(¬R)
∀ ¬(P∧Q)↔ (¬P)∨(¬Q)
If S→ T is a tautology, we write S⇒T.
If S↔ T is a tautology, we write S⇔T.

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A contradiction is a statement that is always
false.
Examples:
• R∧(¬R)
∀ ¬(¬(P∧Q)↔ (¬P)∨(¬Q))
The negation of any tautology is a contradiction, and the negation of any contradiction is
a tautology.
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Exercises
We already know the following tautology:
¬(P∧Q) ⇔ (¬P)∨(¬Q)
Nice home exercise:
Show that ¬(P∨Q) ⇔ (¬P)∧(¬Q).
These two tautologies are known as De
Morgan’s laws.
Table 5 in Section 1.2 shows many useful laws.
Exercises 1 and 7 in Section 1.2 may help you
get used to propositions and operators.
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23

Let’s Talk About Logic
• Logic is a system based on propositions.
• A proposition is a statement that is either
true or false (not both).
• We say that the truth value of a proposition
is either true (T) or false (F).
• Corresponds to 1 and 0 in digital circuits
Fall 2002

CMSC 203 - Discrete

24

Logical Operators (Connectives)
•
•
•
•
•
•

Negation
Conjunction
Disjunction
Exclusive or
Implication
Biconditional

(NOT)
(AND)
(OR)
(XOR)
(if – then)
(if and only if)

Truth tables can be used to show how these
operators can combine propositions to
compound propositions.
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A tautology is a statement that is always true.
Examples:
• R∨(¬R)
∀ ¬(P∧Q)↔ (¬P)∨(¬Q)
If S→ T is a tautology, we write S⇒T.
If S↔ T is a tautology, we write S⇔T.

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A contradiction is a statement that is always
false.
Examples:
• R∧(¬R)
• ¬(¬(P∧Q)↔ (¬P)∨(¬Q))
The negation of any tautology is a contradiction,
and the negation of any contradiction is a
tautology.
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Propositional Functions
Propositional function (open sentence):
statement involving one or more variables,
e.g.: x-3 > 5.
Let us call this propositional function P(x), where
P is the predicate and x is the variable.
What is the truth value of P(2) ?

false


false


true

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Propositional Functions
Let us consider the propositional function
Q(x, y, z) defined as:
x + y = z.
Here, Q is the predicate and x, y, and z are the
variables.
What is the truth value of Q(2, 3, 5) ?

true

What is the truth value of Q(0, 1, 2) ?

false

What is the truth value of Q(9, -9, 0) ?

true

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Universal Quantification
Let P(x) be a propositional function.
Universally quantified sentence:
For all x in the universe of discourse P(x) is true.
Using the universal quantifier ∀:
∀x P(x) “for all x P(x)” or “for every x P(x)”
(Note: ∀x P(x) is either true or false, so it is a
proposition, not a propositional function.)
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Universal Quantification
Example:
S(x): x is a UMBC student.
G(x): x is a genius.
What does ∀x (S(x) → G(x)) mean ?
“If x is a UMBC student, then x is a genius.”
or
“All UMBC students are geniuses.”
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Existential Quantification
Existentially quantified sentence:
There exists an x in the universe of discourse
for which P(x) is true.
Using the existential quantifier ∃:
∃x P(x) “There is an x such that P(x).”
“There is at least one x such that P(x).”
(Note: ∃x P(x) is either true or false, so it is a
proposition, but no propositional function.)
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Existential Quantification
Example:
P(x): x is a UMBC professor.
G(x): x is a genius.
What does ∃x (P(x) ∧ G(x)) mean ?
“There is an x such that x is a UMBC professor
and x is a genius.”
or
“At least one UMBC professor is a genius.”
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Quantification
Another example:
Let the universe of discourse be the real numbers.
What does ∀x∃y (x + y = 320) mean ?
“For every x there exists a y so that x + y = 320.”
Is it true?

yes

Is it true for the natural numbers?

no

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Disproof by Counterexample
A counterexample to ∀x P(x) is an object c so
that P(c) is false.
Statements such as ∀x (P(x) → Q(x)) can be
disproved by simply providing a counterexample.
Statement: “All birds can fly.”
Disproved by counterexample: Penguin.

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Negation
¬(∀x P(x)) is logically equivalent to ∃x (¬P(x)).
¬(∃x P(x)) is logically equivalent to ∀x (¬P(x)).
See Table 3 in Section 1.3.
I recommend exercises 5 and 9 in Section 1.3.

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… and now for something
completely different…

Set Theory
Actually, you will see that logic and
set theory are very closely related.
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Set Theory
• Set: Collection of objects (“elements”)
• a∈A

“a is an element of A”
“a is a member of A”

• a∉A

“a is not an element of A”

• A = {a1, a2, …, an} “A contains…”
• Order of elements is meaningless
• It does not matter how often the same
element is listed.
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Set Equality
Sets A and B are equal if and only if they
contain exactly the same elements.
Examples:
• A = {9, 2, 7, -3}, B = {7, 9, -3, 2} :

A=B

• A = {dog, cat, horse},
B = {cat, horse, squirrel, dog} :

A≠B

• A = {dog, cat, horse},
B = {cat, horse, dog, dog} :

A=B

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Examples for Sets
“Standard” Sets:
• Natural numbers N = {0, 1, 2, 3, …}
• Integers Z = {…, -2, -1, 0, 1, 2, …}
• Positive Integers Z+ = {1, 2, 3, 4, …}
• Real Numbers R = {47.3, -12, π, …}
• Rational Numbers Q = {1.5, 2.6, -3.8, 15, …}
(correct definition will follow)

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Examples for Sets
• A=∅

“empty set/null set”

• A = {z}

Note: z∈A, but z ≠ {z}

• A = {{b, c}, {c, x, d}}
• A = {{x, y}}
Note: {x, y} ∈A, but {x, y} ≠ {{x, y}}
• A = {x | P(x)}
“set of all x such that P(x)”
• A = {x | x∈N ∧ x > 7} = {8, 9, 10, …}
“set builder notation”
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Examples for Sets
We are now able to define the set of rational
numbers Q:
Q = {a/b | a∈Z ∧ b∈Z+}
or
Q = {a/b | a∈Z ∧ b∈Z ∧ b≠0}
And how about the set of real numbers R?
R = {r | r is a real number}
That is the best we can do.
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Subsets
A⊆B
“A is a subset of B”
A ⊆ B if and only if every element of A is also
an element of B.
We can completely formalize this:
A ⊆ B ⇔ ∀x (x∈A → x∈B)
Examples:
A = {3, 9}, B = {5, 9, 1, 3},

A ⊆ B ? true

A = {3, 3, 3, 9}, B = {5, 9, 1, 3}, A ⊆ B ? true
A = {1, 2, 3}, B = {2, 3, 4},
B?
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A⊆

false
43

Subsets

Useful rules:
• A = B ⇔ (A ⊆ B) ∧ (B ⊆ A)
• (A ⊆ B) ∧ (B ⊆ C) ⇒ A ⊆ C (see Venn
Diagram)
U
B

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A

CMSC 203 - Discrete

C

44

Subsets
Useful rules:
∀ ∅ ⊆ A for any set A
• A ⊆ A for any set A
Proper subsets:
A⊂B
“A is a proper subset of B”
A ⊂ B ⇔ ∀x (x∈A → x∈B) ∧ ∃x (x∈B ∧ x∉A)
or
A ⊂ B ⇔ ∀x (x∈A → x∈B) ∧ ¬∀x (x∈B → x∈A)
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Cardinality of Sets
If a set S contains n distinct elements, n∈N,
we call S a finite set with cardinality n.
Examples:
A = {Mercedes, BMW, Porsche}, |A| = 3
B = {1, {2, 3}, {4, 5}, 6}
C=∅
D = { x∈N | x ≤ 7000 }

|B| = 4
|C| = 0
|D| = 7001

E = { x∈N | x ≥ 7000 }

E is infinite!

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The Power Set
P(A)
“power set of A”
P(A) = {B | B ⊆ A}
(contains all subsets of A)
Examples:
A = {x, y, z}
P(A) = {∅ , {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}
A=∅
P(A) = {∅}
Note: |A| = 0, |P(A)| = 1
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The Power Set
Cardinality of power sets:
| P(A) | = 2|A|
• Imagine each element in A has an “on/off” switch
• Each possible switch configuration in A
corresponds to one element in 2A
A
x

1
x

2
x

3
x

4
x

5
x

6
x

7
x

8
x

y
z

y
z

y
z

y
z

y
z

y
z

y
z

y
z

y
z

• For 3 elements in A, there are
2× 2× 2 = 8 elements in P(A)
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Cartesian Product
The ordered n-tuple (a1, a2, a3, …, an) is an ordered
collection of objects.
Two ordered n-tuples (a1, a2, a3, …, an) and
(b1, b2, b3, …, bn) are equal if and only if they
contain exactly the same elements in the same
order, i.e. ai = bi for 1 ≤ i ≤ n.
The Cartesian product of two sets is defined as:
A×B = {(a, b) | a∈A ∧ b∈B}
Example: A = {x, y}, B = {a, b, c}
A×B = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
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Cartesian Product
The Cartesian product of two sets is defined as:
A×B = {(a, b) | a∈A ∧ b∈B}
Example:
A = {good, bad}, B = {student, prof}

}

A×B = { (good, student), (good, prof),

(bad, student), (bad, prof)

B×A = { (student, good), (prof, good),

(student, bad), (prof, bad)

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}

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Cartesian Product
Note that:
• A×∅ = ∅
• ∅×A = ∅
• For non-empty sets A and B: A≠B ⇔ A×B ≠ B×A
• |A×B| = |A|⋅|B|
The Cartesian product of two or more sets is
defined as:
A1×A2×…×An = {(a1, a2, …, an) | ai∈A for 1 ≤ i ≤ n}
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Set Operations
Union: A∪B = {x | x∈A ∨ x∈B}
Example: A = {a, b}, B = {b, c, d}
A∪B = {a, b, c, d}
Intersection: A∩B = {x | x∈A ∧ x∈B}
Example: A = {a, b}, B = {b, c, d}
A∩B = {b}
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Set Operations
Two sets are called disjoint if their intersection
is empty, that is, they share no elements:
A∩B = ∅
The difference between two sets A and B
contains exactly those elements of A that are
not in B:
A-B = {x | x∈A ∧ x∉B}
Example: A = {a, b}, B = {b, c, d}, A-B = {a}

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Set Operations
The complement of a set A contains exactly
those elements under consideration that are not
in A:
Ac = U-A
Example: U = N, B = {250, 251, 252, …}
Bc = {0, 1, 2, …, 248, 249}

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Set Operations
Table 1 in Section 1.5 shows many useful equations.
How can we prove A∪(B∩C) = (A∪B)∩(A∪C)?
Method I:
x∈A∪(B∩C)
⇔ x∈A ∨ x∈(B∩C)
⇔ x∈A ∨ (x∈B ∧ x∈C)
⇔ (x∈A ∨ x∈B) ∧ (x∈A ∨ x∈C)
(distributive law for logical expressions)
⇔ x∈(A∪B) ∧ x∈(A∪C)
⇔ x∈(A∪B)∩(A∪C)
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Set Operations
Method II: Membership table
1 means “x is an element of this set”
0 means “x is not an element of this set”
A B C B∩C

A∪(B∩C)

A ∪B

A ∪C

(A∪B) ∩(A∪C)

0 0 0

0

0

0

0

0

0 0 1

0

0

0

1

0

0 1 0

0

0

1

0

0

0 1 1

1

1

1

1

1

1 0 0

0

1

1

1

1

1 0 1

0

1

1

1

1

1 1 0

0

1

1

1

1

1 1 1

1

1

1

1

1

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Set Operations
Every logical expression can be transformed into an
equivalent expression in set theory and vice versa.
You could work on Exercises 9 and 19 in Section 1.5
to get some practice.

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… and the following mathematical
appetizer is about…

Functions
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Functions
A function f from a set A to a set B is an
assignment of exactly one element of B to each
element of A.
We write
f(a) = b
if b is the unique element of B assigned by the
function f to the element a of A.
If f is a function from A to B, we write
f: A→B
(note: Here, “→“ has nothing to do with if… then)
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Functions
If f:A→B, we say that A is the domain of f and B
is the codomain of f.
If f(a) = b, we say that b is the image of a and a is
the pre-image of b.
The range of f:A→B is the set of all images of
elements of A.
We say that f:A→B maps A to B.

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Functions
Let us take a look at the function f:P→C with
P = {Linda, Max, Kathy, Peter}
C = {Boston, New York, Hong Kong, Moscow}
f(Linda) = Moscow
f(Max) = Boston
f(Kathy) = Hong Kong
f(Peter) = New York
Here, the range of f is C.
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Functions
Let us re-specify f as follows:
f(Linda) = Moscow
f(Max) = Boston
f(Peter) = Boston
Is f still a function? yes
What is its range?
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{Moscow, Boston, Hong Kong}

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Functions
Other ways to represent f:
x

f(x)

Linda

Moscow

Max

Boston
Hong
Kong
Boston

Kathy
Peter
Fall 2002

Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

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Functions
If the domain of our function f is large, it is
convenient to specify f with a formula, e.g.:
f:R→R
f(x) = 2x
This leads to:
f(1) = 2
f(3) = 6
f(-3) = -6
…
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Functions
Let f1 and f2 be functions from A to R.
Then the sum and the product of f1 and f2 are also
functions from A to R defined by:
(f1 + f2)(x) = f1(x) + f2(x)
(f1f2)(x) = f1(x) f2(x)
Example:
f1(x) = 3x, f2(x) = x + 5
(f1 + f2)(x) = f1(x) + f2(x) = 3x + x + 5 = 4x + 5
(f1f2)(x) = f1(x) f2(x) = 3x (x + 5) = 3x2 + 15x
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Functions
We already know that the range of a function
f:A→B is the set of all images of elements a∈A.
If we only regard a subset S⊆A, the set of all
images of elements s∈S is called the image of S.
We denote the image of S by f(S):
f(S) = {f(s) | s∈S}
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Functions
Let us look at the following well-known function:
f(Linda) = Moscow
f(Max) = Boston
f(Peter) = Boston
What is the image of S = {Linda, Max} ?
f(S) = {Moscow, Boston}
What is the image of S = {Max, Peter} ?
f(S) = {Boston}
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Properties of Functions
A function f:A→B is said to be one-to-one (or
injective), if and only if
∀x, y∈A (f(x) = f(y) → x = y)
In other words: f is one-to-one if and only if it
does not map two distinct elements of A onto the
same element of B.

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And again…
f(Linda) = Moscow
f(Max) = Boston
f(Peter) = Boston

g(Linda) = Moscow
g(Max) = Boston
g(Kathy) = Hong Kong
g(Peter) = New York

Is f one-to-one?

Is g one-to-one?

No, Max and Peter are
mapped onto the same
element of the image.

Yes, each element is
assigned a unique
element of the image.

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How can we prove that a function f is one-to-one?
Whenever you want to prove something, first take
a look at the relevant definition(s):
∀x, y∈A (f(x) = f(y) → x = y)
Example:
f:R→R
f(x) = x2
Disproof by counterexample:
f(3) = f(-3), but 3 ≠ -3, so f is not one-to-one.
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… and yet another example:
f:R→R
f(x) = 3x
One-to-one: ∀x, y∈A (f(x) = f(y) → x = y)
To show: f(x) ≠ f(y) whenever x ≠ y
x≠y
⇔ 3x ≠ 3y
⇔ f(x) ≠ f(y),
so if x ≠ y, then f(x) ≠ f(y), that is, f is one-to-one.
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A function f:A→B with A,B ⊆ R is called strictly
increasing, if
∀x,y∈A (x < y → f(x) < f(y)),
and strictly decreasing, if
∀x,y∈A (x < y → f(x) > f(y)).
Obviously, a function that is either strictly
increasing or strictly decreasing is one-to-one.

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A function f:A→B is called onto, or surjective, if
and only if for every element b∈B there is an
element a∈A with f(a) = b.
In other words, f is onto if and only if its range is
its entire codomain.
A function f: A→B is a one-to-one correspondence,
or a bijection, if and only if it is both one-to-one
and onto.
Obviously, if f is a bijection and A and B are finite
sets, then |A| = |B|.
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Examples:
In the following examples, we use the arrow
representation to illustrate functions f:A→B.
In each example, the complete sets A and B are
shown.

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Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

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Is f injective?
No.
Is f surjective?
No.
Is f bijective?
No.

75

Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

Is f injective?
No.
Is f surjective?
Yes.
Is f bijective?
No.

Paul

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Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

Is f injective?
Yes.
Is f surjective?
No.
Is f bijective?
No.

Lübeck
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Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

Is f injective?
No! f is not even
a function!

Lübeck
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Linda

Boston

Max

New York

Kathy

Hong Kong

Peter

Moscow

Helena

Lübeck

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Is f injective?
Yes.
Is f surjective?
Yes.
Is f bijective?
Yes.

79

Inversion
An interesting property of bijections is that
they have an inverse function.
The inverse function of the bijection f:A→B
is the function f-1:B→A with
f-1(b) = a whenever f(a) = b.

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Inversion
Example:

The inverse function
f-1 is given by:

f(Linda) = Moscow
f(Max) = Boston
f(Peter) = Lübeck
f(Helena) = New York

f-1(Moscow) = Linda
f-1(Boston) = Max
f-1(Hong Kong) = Kathy
f-1(Lübeck) = Peter
f-1(New York) = Helena

Clearly, f is bijective.

Inversion is only
possible for bijections
(= invertible functions)

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Inversion
Linda

Boston

f

Max

New York

f-1

Kathy

Hong Kong

Peter

Moscow

Helena

Lübeck

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f-1:C→P is no
function, because
it is not defined
for all elements of
C and assigns two
images to the preimage New York.
82

Composition
The composition of two functions g:A→B and
f:B→C, denoted by f°g, is defined by
(f°g)(a) = f(g(a))
This means that
• first, function g is applied to element a∈A,
mapping it onto an element of B,
• then, function f is applied to this element of
B, mapping it onto an element of C.
• Therefore, the composite function maps
from A to C.
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Composition
Example:
f(x) = 7x – 4, g(x) = 3x,
f:R→R, g:R→R
(f°g)(5) = f(g(5)) = f(15) = 105 – 4 = 101
(f°g)(x) = f(g(x)) = f(3x) = 21x - 4

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Composition
Composition of a function and its inverse:
(f-1°f)(x) = f-1(f(x)) = x
The composition of a function and its inverse
is the identity function i(x) = x.

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Graphs
The graph of a function f:A→B is the set of
ordered pairs {(a, b) | a∈A and f(a) = b}.
The graph is a subset of A×B that can be used
to visualize f in a two-dimensional coordinate
system.

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Floor and Ceiling Functions
The floor and ceiling functions map the real
numbers onto the integers (R→Z).
The floor function assigns to r∈R the largest
z∈Z with z ≤ r, denoted by r.
Examples: 2.3 = 2, 2 = 2, 0.5 = 0, -3.5 = -4
The ceiling function assigns to r∈R the smallest
z∈Z with z ≥ r, denoted by r.
Examples: 2.3 = 3, 2 = 2, 0.5 = 1, -3.5 = -3
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Exercises
I recommend Exercises 1 and 15 in Section 1.6.
It may also be useful to study the graph displays
in that section.
Another question: What do all graph displays for
any function f:R→R have in common?

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… and now for…

Sequences

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Sequences
Sequences represent ordered lists of elements.
A sequence is defined as a function from a subset
of N to a set S. We use the notation an to denote
the image of the integer n. We call an a term of
the sequence.
Example:
subset of N:

1 2 3 4 5 …

S:

2 4 6 8 10 …

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Sequences
We use the notation {an} to describe a sequence.
Important: Do not confuse this with the {} used
in set notation.
It is convenient to describe a sequence with a
formula.
For example, the sequence on the previous slide
can be specified as {an}, where an = 2n.

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The Formula Game
What are the formulas that describe the
following sequences a1, a2, a3, … ?
1, 3, 5, 7, 9, …

an = 2n - 1

-1, 1, -1, 1, -1, …

an = (-1)n

2, 5, 10, 17, 26, …

an = n2 + 1

0.25, 0.5, 0.75, 1, 1.25 … an = 0.25n
3, 9, 27, 81, 243, …
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Strings
Finite sequences are also called strings, denoted
by a1a2a3…an.
The length of a string S is the number of terms
that it consists of.
The empty string contains no terms at all. It has
length zero.

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Summations
What does

n

∑a
j =m

j

stand for?

It represents the sum am + am+1 + am+2 + … + an.
The variable j is called the index of summation,
running from its lower limit m to its upper limit n.
We could as well have used any other letter to
denote this index.

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94

Summations

How can we express the sum of the first 1000
terms of the sequence {an} with an=n2 for
n = 1, 2, 3, … ?
1000
j2 .
We write it as ∑
j =1

What is the value of

6

∑j
j =1

?

It is 1 + 2 + 3 + 4 + 5 + 6 = 21.
What is the value of

100

∑j
j =1

?

It is so much work to calculate this…
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95

Summations
It is said that Friedrich Gauss came up with the
following formula:
n

∑
j =1

n(n + 1)
j=
2

When you have such a formula, the result of any
summation can be calculated much more easily,
for example:
100

∑
j =1

100(100 + 1) 10100
j=
=
= 5050
2
2

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96

Arithemetic Series
How does:

n

∑
j =1

n(n + 1)
j=
2

???

Observe that:
1 + 2 + 3 +…+ n/2 + (n/2 + 1) +…+ (n - 2) + (n - 1) + n
= [1 + n] + [2 + (n - 1)] + [3 + (n - 2)] +…+ [n/2 + (n/2 + 1)]
= (n + 1) + (n + 1) + (n + 1) + … + (n + 1)

(with n/2 terms)

= n(n + 1)/2.

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Geometric Series
How does:

( n +1)

a
−1
∑ a = (a − 1)
j =0
n

j

???

Observe that:
S = 1 + a + a 2 + a3 + … + a n
aS = a + a2 + a3 + … + an + a(n+1)
so, (aS - S) = (a - 1)S = a(n+1) - 1
Therefore, 1 + a + a2 + … + an = (a(n+1) - 1) / (a - 1).
For example: 1 + 2 + 4 + 8 +… + 1024 = 2047.
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Useful Series
1.
2.
3.
4.

n(n + 1)
∑j= 2
j =1
n
a ( n +1) − 1
aj =
∑
(a − 1)
j =0
n
n(n + 1)(2n + 1)
2
∑j =
6
j =1
2
2
n
n (n + 1)
3
∑j = 4
j =1
n

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Double Summations
Corresponding to nested loops in C or Java, there is
also double (or triple etc.) summation:
Example:
5

2

∑∑ ij
i =1 j =1
5

= ∑ (i + 2i )
i =1
5

= ∑ 3i
i =1

= 3 + 6 + 9 + 12 + 15 = 45
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100

Double Summations
Table 2 in Section 1.7 contains some very useful
formulas for calculating sums.
Exercises 15 and 17 make a nice homework.

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101

Enough Mathematical Appetizers!
Let us look at something more interesting:

Algorithms

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102

Algorithms
What is an algorithm?
An algorithm is a finite set of precise instructions
for performing a computation or for solving a
problem.
This is a rather vague definition. You will get to
know a more precise and mathematically useful
definition when you attend CS420.
But this one is good enough for now…
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Algorithms
Properties of algorithms:
•
•
•
•
•
•
•

Input from a specified set,
Output from a specified set (solution),
Definiteness of every step in the computation,
Correctness of output for every possible input,
Finiteness of the number of calculation steps,
Effectiveness of each calculation step and
Generality for a class of problems.
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Algorithm Examples
We will use a pseudocode to specify algorithms,
which slightly reminds us of Basic and Pascal.
Example: an algorithm that finds the maximum
element in a finite sequence
procedure max(a1, a2, …, an: integers)
max := a1
for i := 2 to n
if max < ai then max := ai
{max is the largest element}
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Algorithm Examples
Another example: a linear search algorithm, that
is, an algorithm that linearly searches a sequence
for a particular element.
procedure linear_search(x: integer; a1, a2, …, an:
integers)
i := 1
while (i ≤ n and x ≠ ai)
i := i + 1
if i ≤ n then location := i
else location := 0
{location is the subscript of the term that equals
x, or is zero if x is not found}
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Algorithm Examples
If the terms in a sequence are ordered, a binary
search algorithm is more efficient than linear
search.
The binary search algorithm iteratively restricts
the relevant search interval until it closes in on
the position of the element to be located.

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Algorithm Examples
binary search for the letter ‘j’
search interval
a c d f g h j l m o p r s u v x z
center element

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108

Algorithm Examples
search interval
center element

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109

Algorithm Examples
search interval
center element

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110

Algorithm Examples
search interval
center element

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111

Algorithm Examples
search interval
center element

found !
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112

Algorithm Examples

procedure binary_search(x: integer; a1, a2, …, an:
integers)
i := 1 {i is left endpoint of search interval}
j := n {j is right endpoint of search interval}
while (i < j)
begin
m := (i + j)/2
if x > am then i := m + 1
else j := m
end
if x = ai then location := i
else location := 0
{location is the subscript of the term that equals
x, or is zero if x is not found}
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Complexity
In general, we are not so much interested in the
time and space complexity for small inputs.
For example, while the difference in time
complexity between linear and binary search is
meaningless for a sequence with n = 10, it is
gigantic for n = 230.

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Complexity
For example, let us assume two algorithms A and
B that solve the same class of problems.
The time complexity of A is 5,000n, the one for
B is 1.1n for an input with n elements.
For n = 10, A requires 50,000 steps, but B only 3,
so B seems to be superior to A.
For n = 1000, however, A requires 5,000,000
steps, while B requires 2.5⋅1041 steps.

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Complexity
This means that algorithm B cannot be used for
large inputs, while algorithm A is still feasible.
So what is important is the growth of the
complexity functions.
The growth of time and space complexity with
increasing input size n is a suitable measure for
the comparison of algorithms.

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Complexity
Comparison: time complexity of algorithms A and B
Input Size
n
10
100
1,000
1,000,000
Fall 2002

Algorithm A
5,000n
50,000
500,000
5,000,000
5⋅109

CMSC 203 - Discrete

Algorithm B
1.1n
3
13,781
2.5⋅1041
4.8⋅1041392
117

Complexity
This means that algorithm B cannot be used for
large inputs, while running algorithm A is still
feasible.
So what is important is the growth of the
complexity functions.
The growth of time and space complexity with
increasing input size n is a suitable measure for
the comparison of algorithms.
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118

The Growth of Functions
The growth of functions is usually described
using the big-O notation.
Definition: Let f and g be functions from the
integers or the real numbers to the real numbers.
We say that f(x) is O(g(x)) if there are constants
C and k such that
|f(x)| ≤ C|g(x)|
whenever x > k.
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When we analyze the growth of complexity
functions, f(x) and g(x) are always positive.
Therefore, we can simplify the big-O requirement
to
f(x) ≤ C⋅g(x) whenever x > k.
If we want to show that f(x) is O(g(x)), we only
need to find one pair (C, k) (which is never unique).

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The idea behind the big-O notation is to establish
an upper boundary for the growth of a function
f(x) for large x.
This boundary is specified by a function g(x) that
is usually much simpler than f(x).
We accept the constant C in the requirement
f(x) ≤ C⋅g(x) whenever x > k,
because C does not grow with x.
We are only interested in large x, so it is OK if
f(x) > C⋅g(x) for x ≤ k.
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Example:
Show that f(x) = x2 + 2x + 1 is O(x2).
For x > 1 we have:
x2 + 2x + 1 ≤ x2 + 2x2 + x2
⇒ x2 + 2x + 1 ≤ 4x2
Therefore, for C = 4 and k = 1:
f(x) ≤ Cx2 whenever x > k.
⇒ f(x) is O(x2).
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Question: If f(x) is O(x2), is it also O(x3)?
Yes. x3 grows faster than x2, so x3 grows also
faster than f(x).
Therefore, we always have to find the smallest
simple function g(x) for which f(x) is O(g(x)).

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“Popular” functions g(n) are
n log n, 1, 2n, n2, n!, n, n3, log n
Listed from slowest to fastest growth:
• 1
• log n
• n
• n log n
• n2
• n3
• 2n
• n!
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A problem that can be solved with polynomial
worst-case complexity is called tractable.
Problems of higher complexity are called
intractable.
Problems that no algorithm can solve are called
unsolvable.
You will find out more about this in CS420.
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Useful Rules for Big-O
For any polynomial f(x) = anxn + an-1xn-1 + … + a0, where
a0, a1, …, an are real numbers,
f(x) is O(xn).
If f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then
(f1 + f2)(x) is O(max(g1(x), g2(x)))
If f1(x) is O(g(x)) and f2(x) is O(g(x)), then
(f1 + f2)(x) is O(g(x)).
If f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then
(f1f2)(x) is O(g1(x) g2(x)).
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Complexity Examples
What does the following algorithm compute?
procedure who_knows(a1, a2, …, an: integers)
m := 0
for i := 1 to n-1
for j := i + 1 to n
if |ai – aj| > m then m := |ai – aj|
{m is the maximum difference between any two
numbers in the input sequence}
Comparisons: n-1 + n-2 + n-3 + … + 1
= (n – 1)n/2 = 0.5n2 – 0.5n
Time complexity is O(n2).
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127

Complexity Examples
Another algorithm solving the same problem:
procedure max_diff(a1, a2, …, an: integers)
min := a1
max := a1
for i := 2 to n
if ai < min then min := ai
else if ai > max then max := ai
m := max - min
Comparisons: 2n - 2
Time complexity is O(n).
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Let us get into…

Number Theory

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129

Introduction to Number Theory
Number theory is about integers and their
properties.
We will start with the basic principles of
• divisibility,
• greatest common divisors,
• least common multiples, and
• modular arithmetic
and look at some relevant algorithms.
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Division
If a and b are integers with a ≠ 0, we say that
a divides b if there is an integer c so that b = ac.
When a divides b we say that a is a factor of b
and that b is a multiple of a.
The notation a | b means that a divides b.
We write a X b when a does not divide b
(see book for correct symbol).
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Divisibility Theorems
For integers a, b, and c it is true that
• if a | b and a | c, then a | (b + c)
Example: 3 | 6 and 3 | 9, so 3 | 15.
• if a | b, then a | bc for all integers c
Example: 5 | 10, so 5 | 20, 5 | 30, 5 | 40, …
• if a | b and b | c, then a | c
Example: 4 | 8 and 8 | 24, so 4 | 24.

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Primes
A positive integer p greater than 1 is called prime
if the only positive factors of p are 1 and p.
A positive integer that is greater than 1 and is not
prime is called composite.
The fundamental theorem of arithmetic:
Every positive integer can be written uniquely as
the product of primes, where the prime factors
are written in order of increasing size.
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133

Primes
Examples:
15 =

3·5

48 =

2·2·2·2·3 = 24·3

17 =

17

100 =

2·2·5·5 = 22·52

512 =

2·2·2·2·2·2·2·2·2 = 29

515 =

5·103

28 =

2·2·7

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134

Primes
If n is a composite integer, then n has a prime
divisor less than or equal n .
This is easy to see: if n is a composite integer, it
must have two prime divisors p1 and p2 such that
p1⋅p2 = n.
p1 and p2 cannot both be greater than
n
, because then p1⋅p2 > n.
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The Division Algorithm
Let a be an integer and d a positive integer.
Then there are unique integers q and r, with
0 ≤ r < d, such that a = dq + r.
In the above equation,
• d is called the divisor,
• a is called the dividend,
• q is called the quotient, and
• r is called the remainder.

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Example:
When we divide 17 by 5, we have
17 = 5⋅3 + 2.
•
•
•
•

17 is the dividend,
5 is the divisor,
3 is called the quotient, and
2 is called the remainder.
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Another example:
What happens when we divide -11 by 3 ?
Note that the remainder cannot be negative.
-11 = 3⋅(-4) + 1.
•
•
•
•

-11 is the dividend,
3 is the divisor,
-4 is called the quotient, and
1 is called the remainder.
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138

Greatest Common Divisors
Let a and b be integers, not both zero.
The largest integer d such that d | a and d | b is
called the greatest common divisor of a and b.
The greatest common divisor of a and b is denoted
by gcd(a, b).
Example 1: What is gcd(48, 72) ?
The positive common divisors of 48 and 72 are
1, 2, 3, 4, 6, 8, 12, 16, and 24, so gcd(48, 72) = 24.
Example 2: What is gcd(19, 72) ?
The only positive common divisor of 19 and 72 is
1, so gcd(19, 72) = 1.
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139

Greatest Common Divisors
Using prime factorizations:
a = p1a1 p2a2 … pnan , b = p1b1 p2b2 … pnbn ,
where p1 < p2 < … < pn and ai, bi ∈ N for 1 ≤ i ≤ n
gcd(a, b) = p1min(a1, b1 ) p2min(a2, b2 ) … pnmin(an, bn )
Example:
a = 60 = 22 31 51
b = 54 = 21 33 50
gcd(a, b) = 21 31 50 = 6
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Relatively Prime Integers

Definition:

Two integers a and b are relatively prime if
gcd(a, b) = 1.
Examples:
Are 15 and 28 relatively prime?
Yes, gcd(15, 28) = 1.
Yes, gcd(55, 28) = 1.
No, gcd(35, 28) = 7.
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Relatively Prime Integers
Definition:
The integers a1, a2, …, an are pairwise relatively
prime if gcd(ai, aj) = 1 whenever 1 ≤ i < j ≤ n.
Examples:
Are 15, 17, and 27 pairwise relatively prime?
No, because gcd(15, 27) = 3.
Are 15, 17, and 28 pairwise relatively prime?
Yes, because gcd(15, 17) = 1, gcd(15, 28) = 1 and
gcd(17, 28) = 1.
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Least Common Multiples

Definition:

The least common multiple of the positive
integers a and b is the smallest positive integer
that is divisible by both a and b.
We denote the least common multiple of a and b
by lcm(a, b).
Examples:
lcm(3, 7) = 21
lcm(4, 6) = 12
lcm(5, 10) = 10
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Least Common Multiples
Using prime factorizations:
a = p1a1 p2a2 … pnan , b = p1b1 p2b2 … pnbn ,
where p1 < p2 < … < pn and ai, bi ∈ N for 1 ≤ i ≤ n
lcm(a, b) = p1max(a1, b1 ) p2max(a2, b2 ) … pnmax(an, bn )
Example:
a = 60 = 22 31 51
b = 54 = 21 33 50
lcm(a, b) = 22 33 51 = 4⋅27⋅5 = 540
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GCD and LCM
a = 60 = 22 31 51
b = 54 = 21 33 50
gcd(a, b) =

21 31 50

=6

lcm(a, b) =

22 33 51

= 540

Theorem: a⋅ b = gcd(a,b)⋅ lcm(a,b)
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Modular Arithmetic
Let a be an integer and m be a positive integer.
We denote by a mod m the remainder when a is
divided by m.
Examples:
9 mod 4 = 1
9 mod 3 = 0
9 mod 10 = 9
-13 mod 4 = 3
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146

Congruences
Let a and b be integers and m be a positive integer.
We say that a is congruent to b modulo m if
m divides a – b.
We use the notation a ≡ b (mod m) to indicate
that a is congruent to b modulo m.
In other words:
a ≡ b (mod m) if and only if a mod m = b mod m.

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Congruences
Examples:
Is it true that 46 ≡ 68 (mod 11) ?
Yes, because 11 | (46 – 68).
Is it true that 46 ≡ 68 (mod 22)?
Yes, because 22 | (46 – 68).
For which integers z is it true that z ≡ 12 (mod 10)?
It is true for any z∈{…,-28, -18, -8, 2, 12, 22, 32, …}
Theorem: Let m be a positive integer. The integers
a and b are congruent modulo m if and only if there
is an integer k such that a = b + km.
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Congruences
Theorem: Let m be a positive integer.
If a ≡ b (mod m) and c ≡ d (mod m), then
a + c ≡ b + d (mod m) and ac ≡ bd (mod m).
Proof:
We know that a ≡ b (mod m) and c ≡ d (mod m)
implies that there are integers s and t with
b = a + sm and d = c + tm.
Therefore,
b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and
bd = (a + sm)(c + tm) = ac + m(at + cs + stm).
Hence, a + c ≡ b + d (mod m) and ac ≡ bd (mod m).
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The Euclidean Algorithm
The Euclidean Algorithm finds the greatest
common divisor of two integers a and b.
For example, if we want to find gcd(287, 91), we
divide 287 by 91:
287 = 91⋅3 + 14
We know that for integers a, b and c,
if a | b and a | c, then a | (b + c).
Therefore, any divisor of 287 and 91 must also be
a divisor of 287 - 91⋅3 = 14.
Consequently, gcd(287, 91) = gcd(14, 91).
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In the next step, we divide 91 by 14:
91 = 14⋅6 + 7
This means that gcd(14, 91) = gcd(14, 7).
So we divide 14 by 7:
14 = 7⋅2 + 0
We find that 7 | 14, and thus gcd(14, 7) = 7.
Therefore, gcd(287, 91) = 7.
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In pseudocode, the algorithm can be implemented
as follows:
procedure gcd(a, b: positive integers)
x := a
y := b
while y ≠ 0
begin
r := x mod y
x := y
y := r
end {x is gcd(a, b)}
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Representations of Integers
Let b be a positive integer greater than 1.
Then if n is a positive integer, it can be expressed
uniquely in the form:
n = akbk + ak-1bk-1 + … + a1b + a0,
where k is a nonnegative integer,
a0, a1, …, ak are nonnegative integers less than b,
and ak ≠ 0.
Example for b=10:
859 = 8⋅102 + 5⋅101 + 9⋅100
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Example for b=2 (binary expansion):
(10110)2 = 1⋅24 + 1⋅22 + 1⋅21 = (22)10
Example for b=16 (hexadecimal expansion):
(we use letters A to F to indicate numbers 10 to 15)
(3A0F)16 = 3⋅163 + 10⋅162 + 15⋅160 = (14863)10

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How can we construct the base b expansion of an
integer n?
First, divide n by b to obtain a quotient q0 and
remainder a0, that is,
n = bq0 + a0, where 0 ≤ a0 < b.
The remainder a0 is the rightmost digit in the base b
expansion of n.
Next, divide q0 by b to obtain:
q0 = bq1 + a1, where 0 ≤ a1 < b.
a1 is the second digit from the right in the base b
expansion of n. Continue this process until you
obtain2002
Fall a quotient equal - Discrete
CMSC 203 to zero.
155

Example:
What is the base 8 expansion of (12345)10 ?
First, divide 12345 by 8:
12345 = 8⋅1543 + 1
1543 = 8⋅192 + 7
192 = 8⋅24 + 0
24 = 8⋅3 + 0
3 = 8⋅0 + 3
The result is: (12345)10 = (30071)8.
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procedure base_b_expansion(n, b: positive integers)
q := n
k := 0
while q ≠ 0
begin
ak := q mod b
q := q/b
k := k + 1
end
{the base b expansion of n is (ak-1 … a1a0)b }
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Addition of Integers
Let a = (an-1an-2…a1a0)2, b = (bn-1bn-2…b1b0)2.
How can we add these two binary numbers?
First, add their rightmost bits:
a0 + b0 = c0⋅2 + s0,
where s0 is the rightmost bit in the binary
expansion of a + b, and c0 is the carry.
Then, add the next pair of bits and the carry:
a1 + b1 + c0 = c1⋅2 + s1,
where s1 is the next bit in the binary expansion of
a + b, and c1 is the carry.
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Continue this process until you obtain cn-1.
The leading bit of the sum is sn = cn-1.
The result is:
a + b = (snsn-1…s1s0)2

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Example:
Add a = (1110)2 and b = (1011)2.
a0 + b0 = 0 + 1 = 0⋅2 + 1, so that c0 = 0 and s0 = 1.
a1 + b1 + c0 = 1 + 1 + 0 = 1⋅2 + 0, so c1 = 1 and s1 = 0.
a2 + b2 + c1 = 1 + 0 + 1 = 1⋅2 + 0, so c2 = 1 and s2 = 0.
a3 + b3 + c2 = 1 + 1 + 1 = 1⋅2 + 1, so c3 = 1 and s3 = 1.
s4 = c3 = 1.
Therefore, s = a + b = (11001)2.
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How do we (humans) add two integers?
Example:

1 11
7583
+ 4932

carry

1 25 1 5
Binary expansions:

1 1
(1011)2
+ (1010)2

carry

(101 01 )2
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Let a = (an-1an-2…a1a0)2, b = (bn-1bn-2…b1b0)2.
How can we algorithmically add these two binary
numbers?
First, add their rightmost bits:
a0 + b0 = c0⋅2 + s0,
where s0 is the rightmost bit in the binary
expansion of a + b, and c0 is the carry.
Then, add the next pair of bits and the carry:
a1 + b1 + c0 = c1⋅2 + s1,
where s1 is the next bit in the binary expansion of
162
a +Fall and c1 is CMSC 203 - Discrete
b, 2002
the carry.

Continue this process until you obtain cn-1.
The leading bit of the sum is sn = cn-1.
The result is:
a + b = (snsn-1…s1s0)2

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Example:
Add a = (1110)2 and b = (1011)2.
a0 + b0 = 0 + 1 = 0⋅2 + 1, so that c0 = 0 and s0 = 1.
a1 + b1 + c0 = 1 + 1 + 0 = 1⋅2 + 0, so c1 = 1 and s1 = 0.
a2 + b2 + c1 = 1 + 0 + 1 = 1⋅2 + 0, so c2 = 1 and s2 = 0.
a3 + b3 + c2 = 1 + 1 + 1 = 1⋅2 + 1, so c3 = 1 and s3 = 1.
s4 = c3 = 1.
Therefore, s = a + b = (11001)2.
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procedure add(a, b: positive integers)
c := 0
for j := 0 to n-1
begin
d := (aj + bj + c)/2
sj := aj + bj + c – 2d
c := d
end
sn := c
{the binary expansion of the sum is (snsn-1…s1s0)2}
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Let’s proceed to…

Mathematical
Reasoning

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Mathematical Reasoning
We need mathematical reasoning to
• determine whether a mathematical argument is
correct or incorrect and
• construct mathematical arguments.
Mathematical reasoning is not only important for
conducting proofs and program verification, but
also for artificial intelligence systems (drawing
inferences).

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Terminology
An axiom is a basic assumption about mathematical
structured that needs no proof.
We can use a proof to demonstrate that a
particular statement is true. A proof consists of a
sequence of statements that form an argument.
The steps that connect the statements in such a
sequence are the rules of inference.
Cases of incorrect reasoning are called fallacies.
A theorem is a statement that can be shown to be
true.
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Terminology
A lemma is a simple theorem used as an
intermediate result in the proof of another
theorem.
A corollary is a proposition that follows directly
from a theorem that has been proved.
A conjecture is a statement whose truth value is
unknown. Once it is proven, it becomes a theorem.

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Rules of Inference
Rules of inference provide the justification of
the steps used in a proof.
One important rule is called modus ponens or the
law of detachment. It is based on the tautology
(p∧(p→q)) → q. We write it in the following way:
p
p→q
____
∴q

The two hypotheses p and p → q are
written in a column, and the conclusion
below a bar, where ∴ means “therefore”.

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Rules of Inference
The general form of a rule of inference is:
p1
p2
.
.
.
pn
____
∴q

The rule states that if p1 and p2 and …
and pn are all true, then q is true as well.
These rules of inference can be used in
any mathematical argument and do not
require any proof.

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Rules of Inference
p
_____
Addition
∴ p ∨q

¬q
Modus
p→q
_____ tollens
∴ ¬p

p ∧q
_____
Simplification
∴p

p→q
Hypothetical
q→r
_____ syllogism
∴ p→r

p
q
_____ Conjunction
∴ p ∧q
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p ∨q
Disjunctive
¬p
_____ syllogism
∴q

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Arguments
Just like a rule of inference, an argument consists
of one or more hypotheses and a conclusion.
We say that an argument is valid, if whenever all
its hypotheses are true, its conclusion is also true.
However, if any hypothesis is false, even a valid
argument can lead to an incorrect conclusion.

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Arguments
Example:
“If 101 is divisible by 3, then 1012 is divisible by 9.
101 is divisible by 3. Consequently, 1012 is divisible
by 9.”
Although the argument is valid, its conclusion is
incorrect, because one of the hypotheses is false
(“101 is divisible by 3.”).
If in the above argument we replace 101 with 102,
we could correctly conclude that 1022 is divisible
by 9.
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Arguments

Which rule of inference was used in the last
argument?
p: “101 is divisible by 3.”
q: “1012 is divisible by 9.”
p
Modus
p→q
_____ ponens
∴q
Unfortunately, one of the hypotheses (p) is false.
Therefore, the conclusion q is incorrect.
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Arguments
Another example:
“If it rains today, then we will not have a barbeque
today. If we do not have a barbeque today, then
we will have a barbeque tomorrow.
Therefore, if it rains today, then we will have a
barbeque tomorrow.”
This is a valid argument: If its hypotheses are
true, then its conclusion is also true.
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Arguments
Let us formalize the previous argument:
p: “It is raining today.”
q: “We will not have a barbecue today.”
r: “We will have a barbecue tomorrow.”
So the argument is of the following form:
p→q
Hypothetical
q→r
_____ syllogism
∴ p→r
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Arguments
Another example:
Gary is either intelligent or a good actor.
If Gary is intelligent, then he can count
from 1 to 10.
Gary can only count from 1 to 2.
Therefore, Gary is a good actor.
i: “Gary is intelligent.”
a: “Gary is a good actor.”
c: “Gary can count from 1 to 10.”
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Arguments
i: “Gary is intelligent.”
a: “Gary is a good actor.”
c: “Gary can count from 1 to 10.”
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:

¬c
i→c
¬i
a∨i
a

Hypothesis
Hypothesis
Modus tollens Steps 1 & 2
Hypothesis
Disjunctive Syllogism
Steps 3 & 4

Conclusion: a (“Gary is a good actor.”)
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Arguments
Yet another example:
If you listen to me, you will pass CS 320.
You passed CS 320.
Therefore, you have listened to me.
Is this argument valid?
No, it assumes ((p→q) ∧ q) → p.
This statement is not a tautology. It is false if p is
false and q is true.
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Rules of Inference for Quantified Statements
∀x P(x)
__________
∴ P(c) if c∈U

Universal
instantiation

P(c) for an arbitrary c∈U
___________________
∴ ∀x P(x)

Universal
generalization

∃x P(x)
______________________
∴ P(c) for some element c∈U

Existential
instantiation

P(c) for some element c∈U
____________________
∴ ∃x P(x)

Existential
generalization

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Example:
Every UMB student is a genius.
George is a UMB student.
Therefore, George is a genius.
U(x): “x is a UMB student.”
G(x): “x is a genius.”

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The following steps are used in the argument:
Step 1: ∀x (U(x) → G(x))
Hypothesis
Step 2: U(George) → G(George) Univ. instantiation
using Step 1
Step 3: U(George)
Hypothesis
Step 4: G(George)
Modus ponens
using Steps 2 & 3
∀x P(x)
__________ Universal
∴ P(c) if c∈U instantiation
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Proving Theorems
Direct proof:
An implication p→q can be proved by showing that
if p is true, then q is also true.
Example: Give a direct proof of the theorem
“If n is odd, then n2 is odd.”
Idea: Assume that the hypothesis of this
implication is true (n is odd). Then use rules of
inference and known theorems to show that q
must also be true (n2 is odd).
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Proving Theorems
n is odd.
Then n = 2k + 1, where k is an integer.
Consequently, n2 = (2k + 1)2.
= 4k2 + 4k + 1
= 2(2k2 + 2k) + 1
Since n2 can be written in this form, it is odd.

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Proving Theorems
Indirect proof:
An implication p→q is equivalent to its contrapositive ¬q → ¬p. Therefore, we can prove p→q
by showing that whenever q is false, then p is also
false.
Example: Give an indirect proof of the theorem
“If 3n + 2 is odd, then n is odd.”
Idea: Assume that the conclusion of this
implication is false (n is even). Then use rules of
inference and known theorems to show that p
must also be false (3n + 2 is even).
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Proving Theorems
n is even.
Then n = 2k, where k is an integer.
It follows that 3n + 2 = 3(2k) + 2
= 6k + 2
= 2(3k + 1)
Therefore, 3n + 2 is even.
We have shown that the contrapositive of the
implication is true, so the implication itself is also
true (If 2n + 3 is odd, then n is odd).
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Follow me for a walk through...

Mathematical
Induction

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Induction
The principle of mathematical induction is a
useful tool for proving that a certain predicate
is true for all natural numbers.
It cannot be used to discover theorems, but
only to prove them.

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Induction
If we have a propositional function P(n), and we
want to prove that P(n) is true for any natural
number n, we do the following:
• Show that P(0) is true.
(basis step)
• Show that if P(n) then P(n + 1) for any n∈N.
(inductive step)
• Then P(n) must be true for any n∈N.
(conclusion)
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Induction
Example:
Show that n < 2n for all positive integers n.
Let P(n) be the proposition “n < 2n.”
1. Show that P(1) is true.
(basis step)
P(1) is true, because 1 < 21 = 2.
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Induction
2. Show that if P(n) is true, then P(n + 1) is
true.
(inductive step)
Assume that n < 2n is true.
We need to show that P(n + 1) is true, i.e.
n + 1 < 2n+1
We start from n < 2n:
n + 1 < 2n + 1 ≤ 2n + 2n = 2n+1
Therefore, if n < 2n then n + 1 < 2n+1
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Induction
•

Then P(n) must be true for any positive
integer.
(conclusion)

n < 2n is true for any positive integer.
End of proof.

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Induction
Another Example (“Gauss”):
1 + 2 + … + n = n (n + 1)/2
(basis step)
For n = 0 we get 0 = 0. True.

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Induction
•

Show that if P(n) then P(n + 1) for any n∈N.
(inductive step)

1 + 2 + … + n = n (n + 1)/2
1 + 2 + … + n + (n + 1) = n (n + 1)/2 + (n + 1)
= (2n + 2 + n (n + 1))/2
= (2n + 2 + n2 + n)/2
= (2 + 3n + n2 )/2
= (n + 1) (n + 2)/2
= (n + 1) ((n + 1) + 1)/2
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Induction
•

Then P(n) must be true for any n∈N.
(conclusion)

1 + 2 + … + n = n (n + 1)/2 is true for all n∈N.
End of proof.

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Induction
There is another proof technique that is very
similar to the principle of mathematical induction.
It is called the second principle of
mathematical induction.
It can be used to prove that a propositional
function P(n) is true for any natural number n.

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Induction
The second principle of mathematical induction:
(basis step)
• Show that if P(0) and P(1) and … and P(n),
then P(n + 1) for any n∈N.
(inductive step)
(conclusion)

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Induction
Example: Show that every integer greater than
1 can be written as the product of primes.
(basis step)
2 is the product of one prime: itself.

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Induction
• Show that if P(2) and P(3) and … and P(n),
then P(n + 1) for any n∈N. (inductive step)
Two possible cases:
• If (n + 1) is prime, then obviously P(n + 1) is true.
• If (n + 1) is composite, it can be written as the
product of two integers a and b such that
2 ≤ a ≤ b < n + 1.
By the induction hypothesis, both a and b can be
written as the product of primes.
Therefore, n + 1 = a⋅b can be written as the
product of primes.
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Induction
(conclusion)
End of proof.
We have shown that every integer greater
than 1 can be written as the product of primes.

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If I told you once, it must be...

Recursion
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Recursive Definitions
Recursion is a principle closely related to
mathematical induction.
In a recursive definition, an object is defined in
terms of itself.
We can recursively define sequences, functions
and sets.

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Recursively Defined Sequences
Example:
The sequence {an} of powers of 2 is given by
an = 2n for n = 0, 1, 2, … .
The same sequence can also be defined
recursively:
a0 = 1
an+1 = 2an for n = 0, 1, 2, …
Obviously, induction and recursion are similar
principles.
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Recursively Defined Functions
We can use the following method to define a
function with the natural numbers as its domain:
• Specify the value of the function at zero.
• Give a rule for finding its value at any integer
from its values at smaller integers.
Such a definition is called recursive or inductive
definition.
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Example:
f(0) = 3
f(n + 1) = 2f(n) + 3
f(0) = 3
f(1) = 2f(0) + 3 = 2⋅3 + 3 = 9
f(2) = 2f(1) + 3 = 2⋅9 + 3 = 21
f(3) = 2f(2) + 3 = 2⋅21 + 3 = 45
f(4) = 2f(3) + 3 = 2⋅45 + 3 = 93
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How can we recursively define the factorial
function f(n) = n! ?
f(0) = 1
f(n + 1) = (n + 1)f(n)
f(0) = 1
f(1) = 1f(0) = 1⋅1 = 1
f(2) = 2f(1) = 2⋅1 = 2
f(3) = 3f(2) = 3⋅2 = 6
f(4) = 4f(3) = 4⋅6 = 24
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A famous example: The Fibonacci numbers
f(0) = 0, f(1) = 1
f(n) = f(n – 1) + f(n - 2)
f(0) = 0
f(1) = 1
f(2) = f(1) + f(0) = 1 + 0 = 1
f(3) = f(2) + f(1) = 1 + 1 = 2
f(4) = f(3) + f(2) = 2 + 1 = 3
f(5) = f(4) + f(3) = 3 + 2 = 5
f(6) = f(5) + f(4) = 5 + 3 = 8
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Recursively Defined Sets
If we want to recursively define a set, we need
to provide two things:

• an initial set of elements,
• rules for the construction of additional
elements from elements in the set.
Example: Let S be recursively defined by:
3∈S
(x + y) ∈ S if (x ∈ S) and (y ∈ S)
S is the set of positive integers divisible by 3.
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Proof:
Let A be the set of all positive integers divisible
by 3.
To show that A = S, we must show that
A ⊆ S and S ⊆ A.
Part I: To prove that A ⊆ S, we must show that
every positive integer divisible by 3 is in S.
We will use mathematical induction to show this.
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Let P(n) be the statement “3n belongs to S”.
Basis step: P(1) is true, because 3 is in S.
Inductive step: To show:
If P(n) is true, then P(n + 1) is true.
Assume 3n is in S. Since 3n is in S and 3 is in S, it
follows from the recursive definition of S that
3n + 3 = 3(n + 1) is also in S.
Conclusion of Part I: A ⊆ S.
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Part II: To show: S ⊆ A.
Basis step: To show:
All initial elements of S are in A. 3 is in A. True.
Inductive step: To show:
(x + y) is in A whenever x and y are in S.
If x and y are both in A, it follows that 3 | x and
3 | y. From Theorem I, Section 2.3, it follows
that 3 | (x + y).
Conclusion of Part II: S ⊆ A.
Overall conclusion: A = S.
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Another example:
The well-formed formulae of variables, numerals
and operators from {+, -, *, /, ^} are defined by:
x is a well-formed formula if x is a numeral or
variable.
(f + g), (f – g), (f * g), (f / g), (f ^ g) are wellformed formulae if f and g are.
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With this definition, we can construct formulae
such as:
(x – y)
((z / 3) – y)
((z / 3) – (6 + 5))
((z / (2 * 4)) – (6 + 5))

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Recursive Algorithms
An algorithm is called recursive if it solves a
problem by reducing it to an instance of the same
problem with smaller input.
Example I: Recursive Euclidean Algorithm
procedure gcd(a, b: nonnegative integers with a < b)
if a = 0 then gcd(a, b) := b
else gcd(a, b) := gcd(b mod a, a)

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Example II: Recursive Fibonacci Algorithm
procedure fibo(n: nonnegative integer)
if n = 0 then fibo(0) := 0
else if n = 1 then fibo(1) := 1
else fibo(n) := fibo(n – 1) + fibo(n – 2)

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Recursive Fibonacci Evaluation:
f(4)
f(3)

f(2)

f(2)
f(1)
f(1)
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f(1)

f(0)

f(0)
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procedure iterative_fibo(n: nonnegative integer)
if n = 0 then y := 0
else
begin
x := 0
y := 1
for i := 1 to n-1
begin
z := x + y
x:=y
y := z
end
end {y is the n-th Fibonacci number}
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For every recursive algorithm, there is an
equivalent iterative algorithm.
Recursive algorithms are often shorter, more
elegant, and easier to understand than their
iterative counterparts.
However, iterative algorithms are usually more
efficient in their use of space and time.

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One, two, three, we’re…

Counting
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Basic Counting Principles
Counting problems are of the following kind:
“How many different 8-letter passwords are
there?”
“How many possible ways are there to pick 11
soccer players out of a 20-player team?”
Most importantly, counting is the basis for
computing probabilities of discrete events.
(“What is the probability of winning the lottery?”)
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The sum rule:
If a task can be done in n1 ways and a second task
in n2 ways, and if these two tasks cannot be done
at the same time, then there are n1 + n2 ways to do
either task.
Example:
The department will award a free computer to
either a CS student or a CS professor.
How many different choices are there, if there
are 530 students and 15 professors?
There are 530 + 15 = 545 choices.
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Generalized sum rule:
If we have tasks T1, T2, …, Tm that can be done in
n1, n2, …, nm ways, respectively, and no two of these
tasks can be done at the same time, then there
are n1 + n2 + … + nm ways to do one of these tasks.

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The product rule:
Suppose that a procedure can be broken down
into two successive tasks. If there are n1 ways to
do the first task and n2 ways to do the second
task after the first task has been done, then
there are n1n2 ways to do the procedure.

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Example:
How many different license plates are there that
containing exactly three English letters ?
Solution:
There are 26 possibilities to pick the first letter,
then 26 possibilities for the second one, and 26
for the last one.
So there are 26⋅26⋅26 = 17576 different license
plates.
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Generalized product rule:
If we have a procedure consisting of sequential
tasks T1, T2, …, Tm that can be done in n1, n2, …, nm
ways, respectively, then there are n1 ⋅ n2 ⋅ … ⋅ nm
ways to carry out the procedure.

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The sum and product rules can also be phrased in
terms of set theory.
Sum rule: Let A1, A2, …, Am be disjoint sets. Then
the number of ways to choose any element from
one of these sets is |A1 ∪ A2 ∪ … ∪ Am | =
|A1| + |A2| + … + |Am|.
Product rule: Let A1, A2, …, Am be finite sets. Then
the number of ways to choose one element from
each set in the order A1, A2, …, Am is
|A1 × A2 × … × Am | = |A1| ⋅ |A2| ⋅ … ⋅ |Am|.
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Inclusion-Exclusion
How many bit strings of length 8 either start with a
1 or end with 00?
Task 1: Construct a string of length 8 that starts
with a 1.
There is one way to pick the first bit (1),
two ways to pick the second bit (0 or 1),
two ways to pick the third bit (0 or 1),
.
.
.
two ways to pick the eighth bit (0 or 1).
Product rule: Task 1 can be done in 1⋅27 = 128 ways.
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Inclusion-Exclusion
Task 2: Construct a string of length 8 that ends
with 00.
There are two ways to pick the first bit (0 or 1),
two ways to pick the second bit (0 or 1),
.
.
.
two ways to pick the sixth bit (0 or 1),
one way to pick the seventh bit (0), and
one way to pick the eighth bit (0).
Product rule: Task 2 can be done in 26 = 64 ways.
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Inclusion-Exclusion
Since there are 128 ways to do Task 1 and 64 ways
to do Task 2, does this mean that there are 192 bit
strings either starting with 1 or ending with 00 ?
No, because here Task 1 and Task 2 can be done at
the same time.
When we carry out Task 1 and create strings
starting with 1, some of these strings end with 00.
Therefore, we sometimes do Tasks 1 and 2 at the
same time, so the sum rule does not apply.
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Inclusion-Exclusion
If we want to use the sum rule in such a case, we
have to subtract the cases when Tasks 1 and 2 are
done at the same time.
How many cases are there, that is, how many
strings start with 1 and end with 00?
There is one way to pick the first bit (1),
two ways for the second, …, sixth bit (0 or 1),
one way for the seventh, eighth bit (0).
Product rule: In 25 = 32 cases, Tasks 1 and 2 are
carried out at the same time.
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Inclusion-Exclusion
Since there are 128 ways to complete Task 1 and
64 ways to complete Task 2, and in 32 of these
cases Tasks 1 and 2 are completed at the same
time, there are
128 + 64 – 32 = 160 ways to do either task.
In set theory, this corresponds to sets A1 and A2
that are not disjoint. Then we have:
|A1 ∪ A2| = |A1| + |A2| - |A1 ∩ A2|
This is called the principle of inclusion-exclusion.
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Tree Diagrams
How many bit strings of length four do not have
two consecutive 1s?
Task 1
Task 2
Task 3
Task 4
(1st bit)
(2nd bit)
(3rd bit)
(4th bit)
0

0

0
1
0
0
1
0
1
0

1

0
1
1

0

0
0
1

There are 8 strings.
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The Pigeonhole Principle
The pigeonhole principle: If (k + 1) or more
objects are placed into k boxes, then there is at
least one box containing two or more of the
objects.
Example 1: If there are 11 players in a soccer
team that wins 12-0, there must be at least one
player in the team who scored at least twice.
Example 2: If you have 6 classes from Monday to
Friday, there must be at least one day on which you
have at least two classes.
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The generalized pigeonhole principle: If N
objects are placed into k boxes, then there is at
least one box containing at least N/k of the
objects.
Example 1: In our 60-student class, at least 12
students will get the same letter grade (A, B, C, D,
or F).

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Example 2: Assume you have a drawer containing a
random distribution of a dozen brown socks and a
dozen black socks. It is dark, so how many socks do
you have to pick to be sure that among them there
is a matching pair?
There are two types of socks, so if you pick at
least 3 socks, there must be either at least two
brown socks or at least two black socks.
Generalized pigeonhole principle: 3/2 = 2.
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Permutations and Combinations
How many ways are there to pick a set of 3 people
from a group of 6?
There are 6 choices for the first person, 5 for the
second one, and 4 for the third one, so there are
6⋅5⋅4 = 120 ways to do this.
This is not the correct result!
For example, picking person C, then person A, and
then person E leads to the same group as first
picking E, then C, and then A.
However, these cases are counted separately in
the above equation.
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So how can we compute how many different
subsets of people can be picked (that is, we want to
disregard the order of picking) ?
To find out about this, we need to look at
permutations.
A permutation of a set of distinct objects is an
ordered arrangement of these objects.
An ordered arrangement of r elements of a set is
called an r-permutation.
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Example: Let S = {1, 2, 3}.
The arrangement 3, 1, 2 is a permutation of S.
The arrangement 3, 2 is a 2-permutation of S.
The number of r-permutations of a set with n
distinct elements is denoted by P(n, r).
We can calculate P(n, r) with the product rule:
P(n, r) = n⋅(n – 1)⋅(n – 2) ⋅…⋅(n – r + 1).
(n choices for the first element, (n – 1) for the
second one, (n – 2) for the third one…)
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Example:
P(8, 3) = 8⋅7⋅6 = 336
= (8⋅7⋅6⋅5⋅4⋅3⋅2⋅1)/(5⋅4⋅3⋅2⋅1)
General formula:
P(n, r) = n!/(n – r)!
Knowing this, we can return to our initial question:
from a group of 6 (disregarding the order of
picking)?
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An r-combination of elements of a set is an
unordered selection of r elements from the set.
Thus, an r-combination is simply a subset of the set
with r elements.
Example: Let S = {1, 2, 3, 4}.
Then {1, 3, 4} is a 3-combination from S.
The number of r-combinations of a set with n
distinct elements is denoted by C(n, r).
Example: C(4, 2) = 6, since, for example, the 2combinations of a set {1, 2, 3, 4} are {1, 2}, {1, 3}, {1,
4}, {2, 3}, {2, 4}, {3, 4}.
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How can we calculate C(n, r)?
Consider that we can obtain the r-permutation of a
set in the following way:
First, we form all the r-combinations of the set
(there are C(n, r) such r-combinations).
Then, we generate all possible orderings in each of
these r-combinations (there are P(r, r) such
orderings in each case).
Therefore, we have:
P(n, r) = C(n, r)⋅P(r, r)
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C(n, r) = P(n, r)/P(r, r)
= n!/(n – r)!/(r!/(r – r)!)
= n!/(r!(n – r)!)
Now we can answer our initial question:
from a group of 6 (disregarding the order of
picking)?
C(6, 3) = 6!/(3!⋅3!) = 720/(6⋅6) = 720/36 = 20
There are 20 different ways, that is, 20 different
groups to be picked.
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Corollary:
Let n and r be nonnegative integers with r ≤ n.
Then C(n, r) = C(n, n – r).
Note that “picking a group of r people from a
group of n people” is the same as “splitting a group
of n people into a group of r people and another
group of (n – r) people”.
Please also look at proof on page 252.

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Example:
A soccer club has 8 female and 7 male members.
For today’s match, the coach wants to have 6
female and 5 male players on the grass. How many
possible configurations are there?
C(8, 6) ⋅ C(7, 5) = 8!/(6!⋅2!) ⋅ 7!/(5!⋅2!)
= 28⋅21
= 588

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Combinations
We also saw the following:

n!
n!
C (n, n − r ) =
=
= C (n, r )
(n − r )![n − (n − r )]! (n − r )!r!

This symmetry is intuitively plausible. For example,
let us consider a set containing six elements (n = 6).
Picking two elements and leaving four is essentially
the same as picking four elements and leaving two.
In either case, our number of choices is the
number of possibilities to divide the set into one
set containing two elements and another set
containing four elements.
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Combinations
Pascal’s Identity:
Let n and k be positive integers with n ≥ k.
Then C(n + 1, k) = C(n, k – 1) + C(n, k).
How can this be explained?
What is it good for?

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Combinations
Imagine a set S containing n elements and a set T
containing (n + 1) elements, namely all elements in
S plus a new element a.
Calculating C(n + 1, k) is equivalent to answering
the question: How many subsets of T containing k
items are there?
Case I: The subset contains (k – 1) elements of S
plus the element a: C(n, k – 1) choices.
Case II: The subset contains k elements of S and
does not contain a: C(n, k) choices.
Sum Rule: C(n + 1, k) = C(n, k – 1) + C(n, k).
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Pascal’s Triangle
In Pascal’s triangle, each number is the sum of
the numbers to its upper left and upper right:
1
1
1
1
1
…
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2
3

4
…

1
1
3
6

…

1
4

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…

…
249

Pascal’s Triangle
Since we have C(n + 1, k) = C(n, k – 1) + C(n, k) and
C(0, 0) = 1, we can use Pascal’s triangle to simplify
the computation of C(n, k):
k
C(0, 0) = 1
C(1, 0) = 1 C(1, 1) = 1

n

C(2, 0) = 1 C(2, 1) = 2 C(2, 2) = 1
C(3, 0) = 1 C(3, 1) = 3 C(3, 2) = 3 C(3, 3) = 1
C(4, 0) = 1 C(4, 1) = 4 C(4, 2) = 6 C(4, 3) = 4 C(4, 4) = 1
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Binomial Coefficients
Expressions of the form C(n, k) are also called
binomial coefficients.
How come?
A binomial expression is the sum of two terms,
such as (a + b).
Now consider (a + b)2 = (a + b)(a + b).
When expanding such expressions, we have to
form all possible products of a term in the first
factor and a term in the second factor:
(a + b)2 = a·a + a·b + b·a + b·b
Then we can sum identical terms:
(a + b)2 = a2 + 2ab + b2
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For (a + b)3 = (a + b)(a + b)(a + b) we have
(a + b)3 = aaa + aab + aba + abb + baa + bab + bba + bbb
(a + b)3 = a3 + 3a2b + 3ab2 + b3
There is only one term a3, because there is only one
possibility to form it: Choose a from all three
factors: C(3, 3) = 1.
There is three times the term a2b, because there
are three possibilities to choose a from two out of
the three factors: C(3, 2) = 3.
Similarly, there is three times the term ab2
(C(3, 1) = 3) and once the term b3 (C(3, 0) = 1).
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This leads us to the following formula:
n

( a + b ) n = ∑ C ( n, j ) ⋅ a n − j b j

(Binomial Theorem)

j =0

With the help of Pascal’s triangle, this formula can
considerably simplify the process of expanding powers of
binomial expressions.
For example, the fifth row of Pascal’s triangle
(1 – 4 – 6 – 4 – 1) helps us to compute (a + b)4:
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

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Now it’s Time for…

Recurrence
Relations

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Recurrence Relations
A recurrence relation for the sequence {an} is an
equation that expresses an is terms of one or
more of the previous terms of the sequence,
namely, a0, a1, …, an-1, for all integers n with
n ≥ n0, where n0 is a nonnegative integer.
A sequence is called a solution of a recurrence
relation if it terms satisfy the recurrence
relation.
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In other words, a recurrence relation is like a
recursively defined sequence, but without
specifying any initial values (initial conditions).
Therefore, the same recurrence relation can have
(and usually has) multiple solutions.
If both the initial conditions and the recurrence
relation are specified, then the sequence is
uniquely determined.
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Example:
Consider the recurrence relation
an = 2an-1 – an-2 for n = 2, 3, 4, …
Is the sequence {an} with an=3n a solution of this
recurrence relation?
For n ≥ 2 we see that
2an-1 – an-2 = 2(3(n – 1)) – 3(n – 2) = 3n = an.
Therefore, {an} with an=3n is a solution of the
recurrence relation.
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Is the sequence {an} with an=5 a solution of the
same recurrence relation?
For n ≥ 2 we see that
2an-1 – an-2 = 2⋅5 - 5 = 5 = an.
Therefore, {an} with an=5 is also a solution of the

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Modeling with Recurrence Relations
Example:
Someone deposits $10,000 in a savings account at
a bank yielding 5% per year with interest
compounded annually. How much money will be in
the account after 30 years?
Solution:
Let Pn denote the amount in the account after n
years.
How can we determine Pn on the basis of Pn-1?
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We can derive the following recurrence relation:
Pn = Pn-1 + 0.05Pn-1 = 1.05Pn-1.
The initial condition is P0 = 10,000.
Then we have:
P1 = 1.05P0
P2 = 1.05P1 = (1.05)2P0
P3 = 1.05P2 = (1.05)3P0
…
Pn = 1.05Pn-1 = (1.05)nP0
We now have a formula to calculate Pn for any
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natural number n and can avoid the iteration.

Let us use this formula to find P30 under the
initial condition P0 = 10,000:
P30 = (1.05)30⋅10,000 = 43,219.42

After 30 years, the account contains $43,219.42.

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Another example:
Let an denote the number of bit strings of length
n that do not have two consecutive 0s (“valid
strings”). Find a recurrence relation and give
initial conditions for the sequence {an}.
Solution:
Idea: The number of valid strings equals the
number of valid strings ending with a 0 plus the
number of valid strings ending with a 1.
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Let us assume that n ≥ 3, so that the string
contains at least 3 bits.
Let us further assume that we know the number
an-1 of valid strings of length (n – 1).
Then how many valid strings of length n are there,
if the string ends with a 1?
There are an-1 such strings, namely the set of valid
strings of length (n – 1) with a 1 appended to
them.
Note: Whenever we append a 1 to a valid string,
that string remains valid.
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Now we need to know: How many valid strings of
length n are there, if the string ends with a 0?
Valid strings of length n ending with a 0 must
have a 1 as their (n – 1)st bit (otherwise they
would end with 00 and would not be valid).
And what is the number of valid strings of length
(n – 1) that end with a 1?
We already know that there are an-1 strings of
length n that end with a 1.
Therefore, there are an-2 strings of length (n – 1)
that end with a 1.
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So there are an-2 valid strings of length n that end
with a 0 (all valid strings of length (n – 2) with 10
appended to them).
As we said before, the number of valid strings is
the number of valid strings ending with a 0 plus
the number of valid strings ending with a 1.
That gives us the following recurrence relation:
an = an-1 + an-2
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What are the initial conditions?
a1 = 2 (0 and 1)
a2 = 3 (01, 10, and 11)
a3 = a2 + a1 = 3 + 2 = 5
a4 = a3 + a2 = 5 + 3 = 8
a5 = a4 + a3 = 8 + 5 = 13
…
This sequence satisfies the same recurrence
relation as the Fibonacci sequence.
Since a1 = f3 and a2 = f4, we have an = fn+2.
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Solving Recurrence Relations
In general, we would prefer to have an explicit
formula to compute the value of an rather than
conducting n iterations.
For one class of recurrence relations, we can
obtain such formulas in a systematic way.
Those are the recurrence relations that express
the terms of a sequence as linear combinations of
previous terms.

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Definition: A linear homogeneous recurrence
relation of degree k with constant coefficients is
a recurrence relation of the form:
an = c1an-1 + c2an-2 + … + ckan-k,
Where c1, c2, …, ck are real numbers, and ck ≠ 0.
A sequence satisfying such a recurrence relation
is uniquely determined by the recurrence relation
and the k initial conditions
a0 = C0, a1 = C1, a2 = C2, …, ak-1 = Ck-1.
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Examples:
The recurrence relation Pn = (1.05)Pn-1
is a linear homogeneous recurrence relation of
degree one.
The recurrence relation fn = fn-1 + fn-2
degree two.
The recurrence relation an = an-5
degree five.
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Basically, when solving such recurrence relations,
we try to find solutions of the form an = rn, where
r is a constant.
an = rn is a solution of the recurrence relation
an = c1an-1 + c2an-2 + … + ckan-k if and only if
rn = c1rn-1 + c2rn-2 + … + ckrn-k.
Divide this equation by rn-k and subtract the righthand side from the left:
rk - c1rk-1 - c2rk-2 - … - ck-1r - ck = 0
This is called the characteristic equation of the
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The solutions of this equation are called the
characteristic roots of the recurrence relation.
Let us consider linear homogeneous recurrence
relations of degree two.
Theorem: Let c1 and c2 be real numbers. Suppose
that r2 – c1r – c2 = 0 has two distinct roots r1 and r2.
Then the sequence {an} is a solution of the
recurrence relation an = c1an-1 + c2an-2 if and only if an =
α1r1n + α2r2n for n = 0, 1, 2, …, where α1 and α2 are
constants.
See pp. 321 and 322 for the proof.
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Example: What is the solution of the recurrence
relation an = an-1 + 2an-2 with a0 = 2 and a1 = 7 ?
Solution: The characteristic equation of the
recurrence relation is r2 – r – 2 = 0.
Its roots are r = 2 and r = -1.
Hence, the sequence {an} is a solution to the
recurrence relation if and only if:
an = α12n + α2(-1)n for some constants α1 and α2.
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Given the equation an = α12n + α2(-1)n and the initial
conditions a0 = 2 and a1 = 7, it follows that
a0 = 2 = α1 + α2
a1 = 7 = α1⋅2 + α2 ⋅(-1)
Solving these two equations gives us
α1 = 3 and α2 = -1.
Therefore, the solution to the recurrence relation
and initial conditions is the sequence {an} with
an = 3⋅2n – (-1)n.
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an = rn is a solution of the linear homogeneous
recurrence relation
an = c1an-1 + c2an-2 + … + ckan-k
if and only if
rn = c1rn-1 + c2rn-2 + … + ckrn-k.
Divide this equation by rn-k and subtract the righthand side from the left:
rk - c1rk-1 - c2rk-2 - … - ck-1r - ck = 0
This is called the characteristic equation of the
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The solutions of this equation are called the
characteristic roots of the recurrence relation.
Let us consider linear homogeneous recurrence
relations of degree two.
Theorem: Let c1 and c2 be real numbers. Suppose
that r2 – c1r – c2 = 0 has two distinct roots r1 and r2.
Then the sequence {an} is a solution of the
recurrence relation an = c1an-1 + c2an-2 if and only if an =
α1r1n + α2r2n for n = 0, 1, 2, …, where α1 and α2 are
constants.
See pp. 321 and 322 for the proof.
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Example: Give an explicit formula for the Fibonacci
numbers.
Solution: The Fibonacci numbers satisfy the
recurrence relation fn = fn-1 + fn-2 with initial conditions
f0 = 0 and f1 = 1.
The characteristic equation is r2 – r – 1 = 0.
Its roots are
1+ 5
1− 5
r1 =
, r2 =
2
2
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Therefore, the Fibonacci numbers are given by
n

n

1+ 5 
1− 5 
 + α2

f n = α1 
 2 
 2 




for some constants α1 and α2.
We can determine values for these constants so
that the sequence meets the conditions f0 = 0 and
f1 = 1:
f 0 = α1 + α 2 = 0
1+ 5 
1− 5 
 + α2
 =1
f1 = α1 
 2 
 2 




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The unique solution to this system of two
equations and two variables is
1
1
α1 =
, α2 = −
5
5
So finally we obtained an explicit formula for the
Fibonacci numbers:
n

1 1+ 5 
1 1− 5 

 −


fn =
5 2 
5 2 




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But what happens if the characteristic equation
has only one root?
How can we then match our equation with the initial
conditions a0 and a1 ?
Theorem: Let c1 and c2 be real numbers with c2 ≠ 0.
Suppose that r2 – c1r – c2 = 0 has only one root r0.
A sequence {an} is a solution of the recurrence
relation an = c1an-1 + c2an-2 if and only if
an = α1r0n + α2nr0n, for n = 0, 1, 2, …, where α1 and α2
are constants.
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Example: What is the solution of the recurrence
relation an = 6an-1 – 9an-2 with a0 = 1 and a1 = 6?
Solution: The only root of r2 – 6r + 9 = 0 is r0 = 3.
Hence, the solution to the recurrence relation is
an = α13n + α2n3n for some constants α1 and α2.
To match the initial condition, we need
a0 = 1 = α1
a1 = 6 = α1⋅3 + α2⋅3
Solving these equations yields α1 = 1 and α2 = 1.
Consequently, the overall solution is given by
an = 3n 2002 n. CMSC 203 - Discrete
Fall + n3
280

You Never Escape Your…

Relations
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Relations
If we want to describe a relationship between
elements of two sets A and B, we can use ordered
pairs with their first element taken from A and
their second element taken from B.
Since this is a relation between two sets, it is
called a binary relation.
Definition: Let A and B be sets. A binary relation
from A to B is a subset of A×B.
In other words, for a binary relation R we have
R ⊆ A×B. We use the notation aRb to denote that
(a, b)∈R and aRb to denote that (a, b)∉R.
a
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Relations

When (a, b) belongs to R, a is said to be related to
b by R.
Example: Let P be a set of people, C be a set of
cars, and D be the relation describing which person
drives which car(s).
P = {Carl, Suzanne, Peter, Carla},
C = {Mercedes, BMW, tricycle}
D = {(Carl, Mercedes), (Suzanne, Mercedes),
(Suzanne, BMW), (Peter, tricycle)}
This means that Carl drives a Mercedes, Suzanne
drives a Mercedes and a BMW, Peter drives a
tricycle, and Carla does not drive any of these
vehicles.
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Functions as Relations
You might remember that a function f from a set A
to a set B assigns a unique element of B to each
element of A.
The graph of f is the set of ordered pairs (a, b)
such that b = f(a).
Since the graph of f is a subset of A×B, it is a
relation from A to B.
Moreover, for each element a of A, there is
exactly one ordered pair in the graph that has a as
its first element.
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Functions as Relations
Conversely, if R is a relation from A to B such that
every element in A is the first element of exactly
one ordered pair of R, then a function can be
defined with R as its graph.
This is done by assigning to an element a∈A the
unique element b∈B such that (a, b)∈R.

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Relations on a Set
Definition: A relation on the set A is a relation
from A to A.
In other words, a relation on the set A is a subset
of A×A.
Example: Let A = {1, 2, 3, 4}. Which ordered pairs
are in the relation R = {(a, b) | a < b} ?

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Relations on a Set
Solution: R = { (1, 2), (1, 3), (1, 4), (2, 3),(2, 4),(3, 4)}
1

1

R
1

2

2

3

3

2

3

4

X

X

X

X

X

3

4

1

4

4
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X

287

Relations on a Set

How many different relations can we define on
a set A with n elements?
A relation on a set A is a subset of A×A.
How many elements are in A×A ?
There are n2 elements in A×A, so how many
subsets (= relations on A) does A×A have?
The number of subsets that we can form out of a
m
n2
set with m elements is 2 . Therefore, 2 subsets
can be formed out of A×A.
2

Answer: We can define 2n 2 different relations
on A.
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Properties of Relations

We will now look at some useful ways to classify
relations.
Definition: A relation R on a set A is called
reflexive if (a, a)∈R for every element a∈A.
Are the following relations on {1, 2, 3, 4} reflexive?
R = {(1, 1), (1, 2), (2, 3), (3, 3), (4, 4)}
R = {(1, 1), (2, 2), (2, 3), (3, 3), (4, 4)}
R = {(1, 1), (2, 2), (3, 3)}

No.
Yes.
No.

Definition: A relation on a set A is called
irreflexive if (a, a)∉R for every element a∈A.
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Definitions:
A relation R on a set A is called symmetric if (b,
a)∈R whenever (a, b)∈R for all a, b∈A.
A relation R on a set A is called antisymmetric if
a = b whenever (a, b)∈R and (b, a)∈R.
A relation R on a set A is called asymmetric if
(a, b)∈R implies that (b, a)∉R for all a, b∈A.

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Are the following relations on {1, 2, 3, 4}
symmetric, antisymmetric, or asymmetric?
R = {(1, 1), (1, 2), (2, 1), (3, 3), (4, 4)}
R = {(1, 1)}

symmetric
sym. and
antisym.

R = {(1, 3), (3, 2), (2, 1)}

antisym.
and asym.

R = {(4, 4), (3, 3), (1, 4)}

antisym.

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